Giải thích các bước giải:

Ta có:

\(\begin{array}{l}
18,,\\
3{x^2} - 4x\sqrt {4x - 3}  + 4x - 3 = 0\,\,\,\,\,\,\,\,\,\,\,\left( {x \ge \dfrac{3}{4}} \right)\\
 \Leftrightarrow \left( {3{x^2} - 3x.\sqrt {4x - 3} } \right) - \left( {x.\sqrt {4x - 3}  - \left( {4x - 3} \right)} \right) = 0\\
 \Leftrightarrow 3x\left( {x - \sqrt {4x - 3} } \right) - \sqrt {4x - 3} \left( {x - \sqrt {4x - 3} } \right) = 0\\
 \Leftrightarrow \left( {x - \sqrt {4x - 3} } \right)\left( {3x - \sqrt {4x - 3} } \right) = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
x = \sqrt {4x - 3} \\
3x = \sqrt {4x - 3} 
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
{x^2} = 4x - 3\\
9{x^2} = 4x - 3
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
{x^2} - 4x + 3 = 0\\
9{x^2} - 4x + 3 = 0
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
\left( {x - 1} \right)\left( {x - 3} \right) = 0\\
9{x^2} - 4x + 3 = 0\,\,\,\,\,\,\,\,\,\,\left( {vn} \right)
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
x = 1\\
x = 3
\end{array} \right.\\
19,\\
\sqrt {3x - 2}  + \sqrt {x - 1}  = 4x - 9 + 2\sqrt {3{x^2} - 5x + 2} \,\,\,\,\,\,\,\,\,\,\,\,\,\left( {x \ge 1} \right)\\
 \Leftrightarrow \sqrt {3x - 2}  + \sqrt {x - 1}  = \left( {3x - 2} \right) + 2\sqrt {3{x^2} - 5x + 2}  + \left( {x - 1} \right) - 6\\
 \Leftrightarrow \sqrt {3x - 2}  + \sqrt {x - 1}  = {\sqrt {3x - 2} ^2} + 2.\sqrt {\left( {3x - 2} \right)\left( {x - 1} \right)}  + {\sqrt {x - 1} ^2} - 6\\
 \Leftrightarrow \sqrt {3x - 2}  + \sqrt {x - 1}  = {\left( {\sqrt {3x - 2}  + \sqrt {x - 1} } \right)^2} - 6\\
 \Leftrightarrow {\left( {\sqrt {3x - 2}  + \sqrt {x - 1} } \right)^2} - \left( {\sqrt {3x - 2}  + \sqrt {x - 1} } \right) - 6 = 0\\
 \Leftrightarrow \left[ {\left( {\sqrt {3x - 2}  + \sqrt {x - 1} } \right) - 3} \right].\left[ {\left( {\sqrt {3x - 2}  + \sqrt {x - 1} } \right) + 2} \right] = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
\sqrt {3x - 2}  + \sqrt {x - 1}  = 3\\
\sqrt {3x - 2}  + \sqrt {x - 1}  =  - 2
\end{array} \right.\\
 \Leftrightarrow \sqrt {3x - 2}  + \sqrt {x - 1}  = 3\\
 \Leftrightarrow \left( {\sqrt {3x - 2}  - 2} \right) + \left( {\sqrt {x - 1}  - 1} \right) = 0\\
 \Leftrightarrow \dfrac{{\left( {\sqrt {3x - 2}  - 2} \right)\left( {\sqrt {3x - 2}  + 2} \right)}}{{\sqrt {3x - 2}  + 2}} + \dfrac{{\left( {\sqrt {x - 1}  - 1} \right)\left( {\sqrt {x - 1}  + 1} \right)}}{{\sqrt {x - 1}  + 1}} = 0\\
 \Leftrightarrow \dfrac{{\left( {3x - 2} \right) - {2^2}}}{{\sqrt {3x - 2}  + 2}} + \dfrac{{\left( {x - 1} \right) - {1^2}}}{{\sqrt {x - 1}  + 1}} = 0\\
 \Leftrightarrow \dfrac{{3x - 6}}{{\sqrt {3x - 2}  + 2}} + \dfrac{{x - 2}}{{\sqrt {x - 1}  + 1}} = 0\\
 \Leftrightarrow \left( {x - 2} \right).\left[ {\dfrac{3}{{\sqrt {3x - 2}  + 2}} + \dfrac{1}{{\sqrt {x - 1}  + 1}}} \right] = 0\\
\dfrac{3}{{\sqrt {3x - 2}  + 2}} + \dfrac{1}{{\sqrt {x - 1}  + 1}} > 0,\,\,\,\forall x \ge 1\\
 \Rightarrow x - 2 = 0 \Leftrightarrow x = 2\\
20,\\
\sqrt {2x + 3}  + \sqrt {x - 3}  = 3x - 6 + 2\sqrt {2{x^2} - 3x - 9} \,\,\,\,\,\,\,\,\left( {x \ge 3} \right)\\
 \Leftrightarrow \sqrt {2x + 3}  + \sqrt {x - 3}  = \left( {2x + 3} \right) + 2\sqrt {2{x^2} - 3x - 9}  + \left( {x - 3} \right) - 6\\
 \Leftrightarrow \sqrt {2x + 3}  + \sqrt {x - 3}  = {\sqrt {2x + 3} ^2} + 2.\sqrt {\left( {2x + 3} \right)\left( {x - 3} \right)}  + {\sqrt {x - 3} ^2} - 6\\
 \Leftrightarrow \sqrt {2x + 3}  + \sqrt {x - 3}  = {\left( {\sqrt {2x + 3}  + \sqrt {x - 3} } \right)^2} - 6\\
 \Leftrightarrow {\left( {\sqrt {2x + 3}  + \sqrt {x - 3} } \right)^2} - \left( {\sqrt {2x + 3}  + \sqrt {x - 3} } \right) - 6 = 0\\
 \Leftrightarrow \left[ {\left( {\sqrt {2x + 3}  + \sqrt {x - 3} } \right) - 3} \right].\left[ {\left( {\sqrt {2x + 3}  + \sqrt {x - 3} } \right) + 2} \right] = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
\sqrt {2x + 3}  + \sqrt {x - 3}  = 3\\
\sqrt {2x + 3}  + \sqrt {x - 3}  =  - 2
\end{array} \right.\\
 \Leftrightarrow \sqrt {2x + 3}  + \sqrt {x - 3}  = 3\\
 \Leftrightarrow \left( {\sqrt {2x + 3}  - 3} \right) + \sqrt {x - 3}  = 0\\
 \Leftrightarrow \dfrac{{\left( {\sqrt {2x + 3}  - 3} \right)\left( {\sqrt {2x + 3}  + 3} \right)}}{{\sqrt {2x + 3}  + 3}} + \sqrt {x - 3}  = 0\\
 \Leftrightarrow \dfrac{{\left( {2x + 3} \right) - {3^2}}}{{\sqrt {2x + 3}  + 3}} + \sqrt {x - 3}  = 0\\
 \Leftrightarrow \dfrac{{2\left( {x - 3} \right)}}{{\sqrt {2x + 3}  + 3}} + \sqrt {x - 3}  = 0\\
 \Leftrightarrow \sqrt {x - 3} \left( {\dfrac{{2\sqrt {x - 3} }}{{\sqrt {2x + 3}  + 3}} + 1} \right) = 0\\
 \Rightarrow \dfrac{{2\sqrt {x - 3} }}{{\sqrt {2x + 3}  + 3}} + 1 > 0,\,\,\,\,\forall x\\
 \Rightarrow \sqrt {x - 3}  = 0 \Leftrightarrow x = 3
\end{array}\)